Timeout a Function Call (with Goroutines & Channels)

· 191 words · 1 minute read

Some applications and programs can be very time sensitive - and they often need to return something in a timely fashion. However, it's not always within our control to set a cut off point these operations. Go makes this process somewhat easier though through it's use of goroutines and channels.

In the example below we execute LongRunningProcess which we've given 3 seconds complete - but it contains code to sleep for 5, so it will never complete. We manage this by using a select to listen on multiple channels, one we've created for our function and another one for our timeout.

package main

import (

func main() {

    c1 := make(chan string, 1)

    // Run your long running function in it's own goroutine and pass back it's
    // response into our channel.
    go func() {
        text := LongRunningProcess()
        c1 <- text

    // Listen on our channel AND a timeout channel - which ever happens first.
    select {
    case res := <-c1:
    case <-time.After(3 * time.Second):
        fmt.Println("out of time :(")


func LongRunningProcess() string {
    time.Sleep(5 * time.Second)
    return "My golangcode.com example is finished :)"

timeout a goroutine

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Author:  Edd Turtle

Edd is the Lead Developer at Hoowla, a prop-tech startup, where he spends much of his time working on production-ready Go and PHP code. He loves coding, but also enjoys cycling and camping in his spare time.

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